# Finding weak solutions of conservation law $u_t + (u^4)_x = 0$

Consider the conservation law $$u_t + (u^4)_x = 0,$$ (a) Find the solution at $$t=1$$ with the following initial condition: u(x,0) = \left\lbrace\begin{aligned} &1 && x<0 \\ &2 && 0\leq x \leq 2 \\ &0 && x>2 \end{aligned} \right. . (b) Solve the Riemann problem (You must consider both $$u_l>u_r$$ and $$u_l): u(x,0) = \left\lbrace\begin{aligned} &u_l && x<0 \\ &u_r && x>0 \end{aligned} \right. . (c) Find the Riemann solution at $$x/t = 0$$.

### Try:

The characteristic are given by $$x = 4 g(r)^3 t + r$$ where $$r$$ is parameter. so we have

$$x = \begin{cases} 4t+r, & r<0 \\ 8t+r, & 0 \leq r \leq 2 \\ r, & r > 2 \end{cases}$$

We have two shocks formations at $$x=0$$ and $$x=2$$ for $$t=0$$. We first consider the shock at $$x=0$$, using R=H condition, we want

$$\xi_1'(t) = \frac{ 2^4 - 1^4 }{2-1} = 15 \implies \xi_1(t) = 15t$$

and at $$(x,t) = (2,0)$$ we have

$$\xi_2'(t) = \frac{ - 2^4 }{0-2} = 8 \implies \xi_2(t) = 8t+2$$

So we can write our solution for part a

$$\boxed{ u(x,t) = \begin{cases} 1, & x < 15 t \\ 2, & 15t < x < 8t+2 \\ 0, & x > 8t+2 \end{cases} }$$

IS this correct? I have a question as to what is it that they are asking in c)?

Here is a plot of the characteristic lines in the $$x$$-$$t$$ plane.
Since the flux $$u \mapsto u^4$$ is convex, the classical theory for entropy solutions of conservation laws applies. Without entering into details, the rarefaction wave generated at $$x=0$$ and the shock wave generated at $$x=2$$ leads to the solution u(x,t) = \left\lbrace \begin{aligned} & 1 && x \leq 4 t \\ & \sqrt[3]{x/(4t)} && 4 t \leq x \leq 32 t\\ & 2 && 32 t \leq x \leq 2 + 8 t \\ & 0 && x \geq 2 + 8 t \end{aligned} \right. valid for small times $$t < 1/12$$. For larger times, one must compute the interaction of the rarefaction with the shock (see e.g. related posts on this site). Asking to find the Riemann solution at $$x/t = 0$$ is the same as asking to find the solution at $$x = 0$$ for nonzero time.